Sub-exponential tail bound for Poisson multiplied by cosine of an independent uniform random variable I am looking for the tail bound of the following random variable, hopefully of sub-exponential form:
$\lambda_n^{-1}X_n\cos(\theta_n)$, where $X\sim Poisson(\lambda_n)$ with $\lambda_n\to 0$, and another independent random variable $\theta_n\sim Unif[0,2\pi]$.
So basically I'd like to know how much can that cosine factor "pull" the whole distribution back to its center, therefore improve the concentration result--since $\lambda_n^{-1}X_n$ as sub-exponential$(2\lambda_n^{-1},\lambda_n^{-1})$ random variable is poorly concentrated.
But this is so hard to compute: say I want to compute the MGF. If I first integrate over $\theta_n$, I get the modified Bessel function of the first kind: $$\mathbb{E}e^{\beta\lambda^{-1}X\cos(\theta)}=\mathbb{E}I_0(\frac{\beta}{\lambda}X),$$
which I don't know how to evaluate. Otherwise I get $$\mathbb{E}e^{...}=\mathbb{E}\exp(\lambda(e^{\beta\lambda^{-1}\cos(\theta)}-1)),$$
which is also bizarre.
 A: $\newcommand\la\lambda\newcommand\de\delta\newcommand\si\sigma\newcommand\Th\Theta$There is hardly a reason for having the parameter $n$. So, let $\la:=\la_n\to0$, $X:=X_n$, $\Th:=\theta_n$, $Y:=X/\la$, $Z:=Y\cos\Th$. Note that $EY=1$, $\si_Y:=\sqrt{Var\,Y}=1/\sqrt\la\to\infty$, $EZ=EY\,E\cos\Th=0$, and $\si_Z:=\sqrt{Var\,Z}=\sqrt{EZ^2}=\sqrt{EY^2\,E\cos^2\Th}=\sqrt{EY^2/2}\sim\si_Y/\sqrt2=1/\sqrt{2\la}$. We want to estimate $P(Z-EZ\ge z)=P(Z\ge z)$ for real $z>0$ (note that the distribution of $Z$ is symmetric about $0$). It makes sense to measure the deviation $z$ of $Z$ from its (zero) mean using the standard deviation $\si_Z=1/\sqrt{2\la}$ of $Z$ or, equivalently, $\si_Y=1/\sqrt\la$ as the unit.
For $z>0$ we have
$$P(Z\ge z)\le P(Y\ge z)=P(X\ge\la z)\le P(X>0).$$
Also, for $z\in(0,1/(2\la)]$,
$$P(Z\ge z)\ge P(Y\ge2z)P(\cos\Th\ge1/2)=\tfrac13\,P(Y\ge2z) \\ 
=\tfrac13\,P(X\ge2\la z)=\tfrac13\,P(X>0). $$
So,
$$P(Z\ge z)\asymp P(X>0)\sim\la\to0$$
for $z\in(0,1/(2\la)]$. Note also that $1/(2\la)$ is much greater than the standard deviation $\si_Z\sim1/\sqrt{2\la}$ of $Z$.

Similarly, for each $k\in\{0,1,\dots\}$ and each $\de\in(0,1)$,
$$P(Z\ge z)\asymp\frac{\la^{k+1}}{(k+1)!} \tag{1}\label{1}$$
for $z\in(k/\la,(k+1-\de)/\la]$.
Note also that for all real $z>0$
$$P(Z\ge z)=2e^{-\la}\sum_{j\ge\la z}\frac{\la^j}{j!}\arccos\frac{\la z}j.$$

The latter paragraph is illustrated below by the plots $\Big\{P(Z\ge z)\Big/\frac{\la^{k+1}}{(k+1)!}\colon z\in(k/\la,(k+1-\de)/\la]\Big\}$
of the ratios of the left-hand side of \eqref{1} to its right-hand side for $k=3$, $\de=0.1$, $\la=0.1$ (red), $\la=0.03$ (green), and $\la=0.01$ (blue):

